Python：这是否可以覆盖__eq__和__hash__？

Question

我是Python的新手，我想确保我正确地覆盖__eq__和__hash__，以免以后造成痛苦的错误：

（我正在使用Google App Engine。）

class Course(db.Model):
    dept_code = db.StringProperty()
    number = db.IntegerProperty()
    title = db.StringProperty()
    raw_pre_reqs = db.StringProperty(multiline=True)
    original_description = db.StringProperty()

    def getPreReqs(self):
        return pickle.loads(str(self.raw_pre_reqs))

    def __repr__(self):
        title_msg = self.title if self.title else "Untitled"
        return "%s %s: %s" % (self.dept_code, self.number, title_msg)

    def __attrs(self):
        return (self.dept_code, self.number, self.title, self.raw_pre_reqs, self.original_description)

    def __eq__(self, other):
        return isinstance(other, Course) and self.__attrs() == other.__attrs()

    def __hash__(self):
        return hash(self.__attrs())

稍微复杂的类型：

class DependencyArcTail(db.Model):
    ''' A list of courses that is a pre-req for something else '''
    courses = db.ListProperty(db.Key)

    ''' a list of heads that reference this one '''
    forwardLinks = db.ListProperty(db.Key)

    def __repr__(self):
        return "DepArcTail %d: courses='%s' forwardLinks='%s'" % (id(self), getReprOfKeys(self.courses), getIdOfKeys(self.forwardLinks))

    def __eq__(self, other):
        if not isinstance(other, DependencyArcTail):
            return False

        for this_course in self.courses:
            if not (this_course in other.courses):
                return False

        for other_course in other.courses:
            if not (other_course in self.courses):
                return False

        return True

    def __hash__(self):
        return hash((tuple(self.courses), tuple(self.forwardLinks)))

一切都很好看？

更新以反映@Alex的评论

class DependencyArcTail(db.Model):
    ''' A list of courses that is a pre-req for something else '''
    courses = db.ListProperty(db.Key)

    ''' a list of heads that reference this one '''
    forwardLinks = db.ListProperty(db.Key)

    def __repr__(self):
        return "DepArcTail %d: courses='%s' forwardLinks='%s'" % (id(self), getReprOfKeys(self.courses), getIdOfKeys(self.forwardLinks))

    def __eq__(self, other):
        return isinstance(other, DependencyArcTail) and set(self.courses) == set(other.courses) and set(self.forwardLinks) == set(other.forwardLinks)

    def __hash__(self):
        return hash((tuple(self.courses), tuple(self.forwardLinks)))

Answer 1

第一个很好。第二个问题有两个原因：

1. .courses
中可能存在重复项
2. 具有相同.courses但不同.forwardLinks的两个实体将比较相等但具有不同的哈希值

我会通过使等式依赖于课程和前向链接来修复第二个，但两者都改变​​为集合（因此没有重复），并且对于散列也是如此。即：

def __eq__(self, other):
    if not isinstance(other, DependencyArcTail):
        return False

    return (set(self.courses) == set(other.courses) and
            set(self.forwardLinks) == set(other.forwardLinks))

def __hash__(self):
    return hash((frozenset(self.courses), frozenset(self.forwardLinks)))

这当然假设前向链接 对于对象的“实际值”至关重要，否则应该从__eq__和__hash__中省略它们。

修改：从对__hash__ tuple的{​​{1}}次电话中移除，这些电话最多是冗余的（可能会受到损害，正如@Mark评论所建议的那样[[tx !!! ]]）;根据@Phillips [[tx !!!]]的评论建议，在散列中将set更改为frozenset。